A compact form of expressing lengthy sums is the use of summation notation which is often known as sigma notation because it uses the Greek letter $\Sigma$ (uppercase sigma, corresponding to English letter “S” that stands for sum), to represent the sum.

To show how this notation works, consider the sum
$1^{2}+2^{2}+3^{2}+4^{2}+\cdots+100^{2}.$ A typical term in this sum is of the form $k^{2}$ and we can get all the terms if we let $k$ run through the values $1, 2, 3,\dots , 100$. In sigma notation, this sum will be written as
$\sum_{k=1}^{100}k^2.$ This symbol is read “the summation of $k^{2}$ where $k$ runs from 1 to 100.”

In general, if $m$ and $n$ are two integers such that $m\le n$, and $f(k)$ is some formula in $k$,
$\sum_{k=m}^{n}f(k)$ denotes the sum of all the terms that we get by substituting integers for $k$ in $f(k)$ starting with $k=m$ and ending with $k=n$. That is,
$\sum_{k=m}^{n}f(k)=f(m)+f(m+1)+f(m+2)+\cdots+f(n-1)+f(n).$ The numbers $m$ and $n$ that appear under and above the sigma are, respectively, called the lower and upper limits of summation, the letter $k$ is called the index of summation.

•  Of course, any convenient letter that is not reserved for another purpose can be used in place of k. For example,
$\sum_{i=1}^{100}i^{2},\qquad\sum_{j=1}^{100}j^{2},\qquad\sum_{j=1}^{100}m^{2},\qquad\sum_{n=1}^{100}n^{2}$ all denote the same sum $\sum_{k=1}^{100}k^{2}$. The letters $i,j,k,m,n$, etc. that are used as the index of summation are called dummy indices.

Here are some examples of using summation notation:

1. ${\displaystyle \sum_{k=1}^{4}k^{3}=1^{3}+2^{3}+3^{3}+4^{3}}$
2. ${\displaystyle \sum_{i=1}^{5}(2i-1)=1+3+5+7+9}$
3. ${\displaystyle \sum_{n=-3}^{4}2^{n}=2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3}+2^{4}}$
4. ${\displaystyle \sum_{k=1}^{n}1=\underbrace{1+1+\cdots+1}_{n\text{ terms}}=n}$

• We can change the upper and lower limits of the sigma notation if we suitably change the formula of the typical term. For example, it is easy to see
$\sum_{n=2}^{5}n^{4}=\sum_{k=0}^{3}(k+2)^{4}=\sum_{m=1}^{4}(m+1)^{4}=2^{4}+3^{4}+4^{4}+5^{4}$

To state general properties of sums, in place of of the notations $f(k),a(k),$ and $b(k)$, representing different formulas in $k$, it is a convention to use a subscripted letter and write $f_{k},a_{k}$, and $b_{k}.$ For example, if $a_{k}=2k$ then
\begin{align*}
\sum_{k=1}^{5}a_{k} & =a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\\
& =2\cdot1+2\cdot2+2\cdot3+2\cdot4+2\cdot5
\end{align*}

For manipulating sums, the following properties of the sigma notation come in very handy.

1. Additive property: Sigma distributes across sums
${\displaystyle \sum_{k=m}^{n}(a_{k}+b_{k})=\sum_{k=m}^{n}a_{k}+\sum_{k=m}^{n}b_{k}}$
2. Homogeneous property: A constant can be moved through a sigma sum:
${\displaystyle \sum_{k=m}^{n}ca_{k}=c\sum_{k=m}^{n}a_{k}}$ where $c$ does not depend on $k$

3. If $m\leq n$ and $p+1\leq n$ then $\sum_{k=m}^na_k=\sum_{k=m}^p a_k+\sum_{k=p+1}^n a_k$
4. If $a_k\leq b_k$ for all $k$ with $m\leq k\leq m$ then
$\sum_{k=m}^n a_k\leq \sum_{k=m}^n b_k.$
5. Telescoping property
${\displaystyle \sum_{k=m}^{n}(a_{k}-a_{k-1})=a_{n}-a_{m-1}}$

#### Proofs of Properties

To prove the above properties, we just need to expand both sides and use properties of real numbers. For property (1), we write out the left hand side. Then because addition is associative and commutative we can rearrange the terms as the right hand side:
$(a_{m}+b_{m})+(a_{m+1}+b_{m+1})+\cdots+(a_{n}+b_{n})=(a_{m}+a_{m+1}+\cdots+a_{m})+(b_{m}+b_{m+1}\cdots+b_{n})$ Property (2) follows from the distributive property of real numbers:
$ca_{m}+ca_{m+1}+\cdots+ca_{n}=c(a_{m}+a_{m+1}+\cdots+a_{n}).$ Property (3) says
$a_m+a_{m+1}+\cdots+a_n=(a_m+\cdots+a_p)+(a_{p+1}+\cdots+a_n),$ which is a generalization of the associative law.

Property (4) is a generalization of the basic law of inequalities:
$a_1\leq b_1,\quad\text{and}\quad a_2\leq b_2\quad\Rightarrow\quad a_1+a_2\leq b_1+b_2.$

For property (5):
\begin{align*}
\sum_{k=m}^{n}(a_{k}-a_{k-1}) & =(\cancel{a_{m}}-a_{m-1})+(\bcancel{a_{m+1}}-\cancel{a_{m}})+\cdots+(a_{n}-\xcancel{a_{n-1}})\\
& =a_{n}-a_{m-1}
\end{align*}

It follows from (1) and (2) that
\begin{align*}
\sum_{k=m}^{n}(ca_{k}+db_{k}) & =\sum_{k=m}^{n}ca_{k}+\sum_{k=m}^{n}db_{k}\\
& =c\sum_{k=m}^{n}a_{k}+d\sum_{k=m}^{n}b_{k}
\end{align*}
and
\begin{align*}
\sum_{k=m}^{n}(a_{k}-b_{k}) & =\sum_{k=m}^{n}(a_{k}+(-1)b_{k})\\
& =\sum_{k=m}^{n}a_{k}+(-1)\sum_{k=m}^{n}b_{k}\\
& =\sum_{k=m}^{n}a_{k}-\sum_{k=m}^{n}b_{k}
\end{align*}

In general
$\sum_{k=1}^{n}(a_{k}b_{k})\neq\left(\sum_{k=1}^{n}a_{k}\right)\left(\sum_{k=1}^{n}b_{k}\right)\qquad\sum_{k=1}^{n}\frac{a_{k}}{b_{k}}\neq\frac{\sum_{k=1}^{n}a_{k}}{\sum_{k=1}^{n}b_{k}}$

Here are some important formulas that are useful in calculus

$\bbox[#F2F2F2,5px,border:2px solid black]{\large\sum_{k=1}^{n}k=1+2+\cdots+n=\frac{n(n+1)}{2}} \quad \text{(i)}$

$\bbox[#F2F2F2,5px,border:2px solid black]{\large \sum_{k=1}^{n}k^{2}=1^{2}+2^{2}+\cdots+n^{2}=\frac{n(n+1)(2n+1)}{6}} \quad \text{(ii)}$

$\bbox[#F2F2F2,5px,border:2px solid black]{\large\sum_{k=1}^{n}k^{3}=\left[\frac{n(n+1)}{2}\right]^{2}} \quad \text{(iii)}$

There are various ways to prove the above formulas. For example, we can use mathematical induction (see Wikipedia Page on mathematical induction ) or use the telescoping property of sigma notation.

#### Proofs of Formulas (i)–(iii)

$\sum_{k=1}^{n}\left(k^{2}-(k-1)^{2}\right)=n^{2}-0^{2}=n^{2}$ But $2k-1=k^{2}-(k-1)^{2}$. Therefore

$\therefore\sum_{k=1}^{n}(2k-1)=n^{2}$

\begin{align*}
\Rightarrow \sum_{k=1}^n 2k-\sum_{k=1}^n 1 &=n^2 &{\small (\text{Property 1})}\\
2\sum_{k=1}^n k-n&=n^2 &{\small (\text{Property 2 and }\sum_{k=1}^n 1=n)}
\end{align*}

Thus we obtain Formula (i)
$\sum_{k=1}^{n}k=\frac{n^{2}+n}{2}=\frac{n(n+1)}{2}$

$\sum_{k=1}^{n}\left(k^{3}-(k-1)^{3}\right)=n^{3}-0^{3}=n^{3}$ But $k^{3}-(k-1)^{3}=3k^{2}-3k+1$ [to see how we obtain it, just expand $(k-1)^{3}$]. Therefore,

\begin{align*}
\therefore\sum_{k=1}^{n}\left(3k^{2}-3k+1\right) & =n^{3}\\
3\sum_{k=1}^{n}k^{2}-3\sum_{k=1}^{n}k+\sum_{k=1}^{n}1 & =n^{3}& {\small(\text{Properties 1 and 2})}
\end{align*}
Now we use Formula (i) to replace $\sum_{k=1}^{n}k$ and the fact
that $\sum_{k=1}^{n}1=n$ to obtain

$3\sum_{k=1}^{n}k^{2}-3\frac{n(n+1)}{2}+n=n^{3}$ After simplification, we get
$\sum_{k=1}^{n}k^{2}=\frac{n^{3}}{3}+\frac{n^{2}}{2}+\frac{n}{6}.$ Using the least common denominator and then factorization, we can show
$\frac{n^{3}}{3}+\frac{n^{2}}{2}+\frac{n}{6}=\frac{n(n+1)(2n+1)}{6}$ Thus
$\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$

$\sum_{k=1}^{n}\left(k^{4}-(k-1)^{4}\right)=n^{4}-0^{4}=n^{4}$ and $k^{4}-(k-1)^{4}=4k^{3}-6k^{2}+4k-1$.
\begin{align*}
\therefore\sum_{k=1}^{n}(4k^{3}-6k^{2}+4k-1) & =n^{4}\\
4\sum_{k=1}^{n}k^{3}-6\sum_{k=1}^{n}k^{2}+4\sum_{k=1}^{n}k-\sum_{k=1}^{n}1 & =n^{4}
\end{align*}
If we use Formulas (i) and (ii), the fact that $\sum_{k=1}^{n}1=n$, and simplify, we obtain
$\sum_{k=1}^{n}k^{3}=\frac{n^{2}(n^{2}+2n+1)}{4}=\left[\frac{n(n+1)}{2}\right]^{2}.$

Example 1
Evaluate ${\displaystyle \sum_{k=1}^{45}k(k+1)}$
Solution
\begin{align*}
{\displaystyle \sum_{k=1}^{45}k(k+1)} & ={\displaystyle \sum_{k=1}^{45}(k^{2}+k)}\\
& ={\displaystyle \sum_{k=1}^{45}k^{2}}+{\displaystyle \sum_{k=1}^{45}k}\\
& =\frac{45\times46\times(2\times45+1)}{6}+\frac{45\times46}{2}\\
& =31395+1035\\
& =32430.
\end{align*}
Example 2
Evaluate ${\displaystyle \sum_{k=5}^{30}\frac{k^{3}}{7}}$
Solution
\begin{align*}
\sum_{k=5}^{30}\frac{k^{3}}{7} & =\frac{1}{7}\sum_{k=5}^{30}k^{3}\\
& =\frac{1}{7}\left(\sum_{k=1}^{30}k^{3}-\sum_{k=1}^{4}k^{3}\right)\\
& =\frac{1}{7}\left(\left[\frac{30\times31}{2}\right]^{2}-\left[\frac{4\times5}{2}\right]^{2}\right)\\
& =\frac{1}{7}\left(465^{2}-10^{2}\right)\\
& =\frac{216125}{7}=30875.
\end{align*}
Example 3
Express ${\displaystyle \sum_{k=1}^{n}(2+k)^{2}}$ in closed form.
Solution
\begin{align*}
\sum_{k=1}^{n}(2+k)^{2} & =\sum_{k=1}^{n}(4+4k+k^{2}) &{\small (\text{Expand} (2+k)^2)}\\
& =4\sum_{k=1}^{n}1+4\sum_{k=1}^{n}k+\sum_{k=1}^{n}k^{2} &{\small  (\text {Properties (1) and (2)})}\\
& =4n+4\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6} & {\small (\text{Formulas (i) and (ii)})}\\
& =4n+(2n^{2}+2n)+\left(\frac{n^{3}}{3}+\frac{n^{2}}{2}+\frac{n}{6}\right)\\
& =\frac{n^{3}}{3}+\frac{5n^{2}}{2}+\frac{37n}{6}\\
& =\frac{2n^{3}+15n^{2}+37n}{6}.
\end{align*}